Maths · 10 Statistics

10.7 Scatter diagrams

Before you start

Plotting and reading points $(x, y)$ on a grid whose scale is not one unit per square.
Working out the mean of a list: add the values and divide by how many there are.
Finding the gradient of a straight line from two points on it, and reading $m$ and $c$ in $y = m x + c$ .
Substituting a value into a formula such as $y = 2.5 x + 7$ .
Finding the statistics (list) mode on your graphic display calculator. You need it in the last section only.

Syllabus objectives covered (4)

Draw and interpret scatter diagrams.
Understand what is meant by positive, negative and zero correlation.
Draw by eye, interpret and use a straight line of best fit.
Use a graphic display calculator to find and use the equation of linear regression.

Part 1 of 6

What a scatter diagram shows, and how to draw one

A scatter diagram compares two numerical variables recorded for the same items. For each item you have a pair of values: the noon temperature on a day and the number of hot drinks a café sold that day, or the age of a bicycle and its price.
Each pair becomes one point $(x, y)$ . Eight days of café records give eight points. The points are the whole diagram: you do not join them up with lines.
Put the variable you would use to make a prediction on the horizontal axis. The café owner wants to predict sales from the weather forecast, so temperature goes across and drinks sold goes up. If a question gives you a grid with labelled axes, use the axes as given.
Choose a scale for each axis that stays the same all the way along and fits all the data on the grid. An axis does not have to start at $0$ : if every number of drinks is between $40$ and $100$ , the vertical axis can run from $40$ to $100$ . Label each axis with the variable and its unit.
Before you plot anything, work out what one small square is worth on each axis. If $10$ small squares stand for $5$ °C, one small square is $5 \div 10 = 0.5$ °C. A wrong value here moves every point, so do it first.
Mark each point with a small cross ×, with the centre of the cross exactly on the point. Plot every pair, then check each cross against the table a second time: it is easy to read across to the wrong row.

Part 2 of 6

Correlation: the pattern in the points

Correlation describes how two variables change together. You judge it from the overall pattern of all the points, not from one or two of them.
Positive correlation: as one variable increases, the other tends to increase. The points rise from bottom left to top right. A person's height and their arm span show positive correlation: taller people tend to have a longer arm span.
Negative correlation: as one variable increases, the other tends to decrease. The points fall from top left to bottom right. The café data later in this lesson show this: on hotter days, fewer hot drinks are sold.
Zero correlation (also called no correlation): the points are scattered with no upward or downward trend. Knowing one variable tells you nothing about the other. An adult's height and the day of the month on which they were born show zero correlation.
graph diagram — not drawn yet
Three small scatter diagrams side by side, each with about twelve crosses on plain x and y axes: the first, labelled 'positive', has crosses rising from bottom left to top right close to a straight line; the second, labelled 'negative', has crosses falling from top left to bottom right; the third, labelled 'zero', has crosses spread over the whole grid with no upward or downward trend.
Look at the direction of the overall trend: up to the right is positive, down to the right is negative, no direction at all is zero correlation.
The word 'tends' matters. In a positive correlation some points sit lower than a point to their left. That does not cancel the correlation: the overall trend is what counts.
Correlation can also be strong or weak. Strong: the points lie close to a straight line. Weak: the points show the trend but are spread widely around it.
The trap: 'negative' is a direction, not a strength. A negative correlation can be strong. If the points show no trend at all, the word is 'zero', not 'negative'.

Part 3 of 6

Interpreting a scatter diagram in context

Naming the correlation is not the same as interpreting it. 'Negative correlation' names the pattern. 'On hotter days the café tends to sell fewer hot drinks' says what the pattern means for the things in the question.
When a question asks what the diagram shows about the relationship, write a sentence in context. Use the variables' own names and the word 'tends': 'the older the bicycle, the lower its price tends to be'. That gives the direction and admits that not every point fits.
An outlier is a point well away from the pattern of the others. It may be a recording error or an unusual item, such as a day when the café had a large group booking. If a question asks for it, give its coordinates.
A correlation does not, by itself, show that one variable causes the other to change. Sales of cold drinks and the number of people at a beach may both rise in hot weather. They show positive correlation, but buying a drink does not send people to the beach: the weather affects both.

Part 4 of 6

Drawing a line of best fit by eye

When the points show positive or negative correlation, you can draw a line of best fit: one straight line, drawn with a ruler, that follows the trend. With zero correlation there is no trend to follow, so there is no line of best fit.
The line must pass through the mean point $(\overline{x}, \overline{y})$ . Here $\overline{x}$ is the mean of the $x$ -values and $\overline{y}$ is the mean of the $y$ -values, for $n$ pairs. So calculate the mean point first.
$\overline{x} = \frac{sum of the x -values}{n}, \overline{y} = \frac{sum of the y -values}{n}$
Plot the mean point and mark it clearly, for example with a small circle around it, so that you do not mistake it for one of the data points.
Lay your ruler through the mean point and turn it about that point until the points balance: roughly as many above the line as below it, with points on both sides along its whole length. If all the points at one end are above the line and all the points at the other end are below it, the angle is wrong, even if the totals balance.
Draw the line across the full data set, from the smallest $x$ -value to the largest. A short line through the middle does not show the trend of all the data.
The line does not have to pass through any of the data points. It does not have to pass through the origin. It is not the line joining the first and last points: those are only two of the data, and either one could be an outlier.
graph diagram — not drawn yet
Two copies of the same scatter diagram of ten crosses showing positive correlation, each with the mean point marked by a small circle. Left copy, labelled 'good': one ruled line through the circled mean point, running from the smallest to the largest x-value, with crosses on both sides of it along its whole length. Right copy, labelled 'poor': a ruled line joining the leftmost cross to the rightmost cross, which misses the circled mean point and has most of the other crosses below it.
The good line goes through the mean point and balances the points along its whole length. The line joining the two end points ignores the other eight points.

Part 5 of 6

Using a line of best fit

To estimate $y$ for a given $x$ : go up from the $x$ -value to the line, then across to the $y$ -axis, and read the value. Draw these reading lines lightly on the graph so that your method can be seen.
To estimate $x$ for a given $y$ : go across from the $y$ -value to the line, then down to the $x$ -axis.
Your estimate depends on your line, so two good lines give slightly different answers. That is expected for a line drawn by eye.
An estimate inside the range of the data is called interpolation. It is reasonably reliable, and more so when the correlation is strong.
An estimate outside the range of the data is called extrapolation. It is unreliable: you have no data there, so you do not know that the pattern continues. Extend the café line in Worked example 2 to $45$ °C and it predicts a negative number of hot drinks, which is impossible.
The gradient of the line tells you how much $y$ changes, on average, for each increase of $1$ in $x$ . Find it from two points on your line that are far apart, not from two data points, because the data points are not on the line.
$gradient = \frac{change in y}{change in x}$
Interpret the gradient in context, with units. A gradient of $- 2.8$ for the café line means that about $2.8$ fewer hot drinks are sold for each $1$ °C rise in temperature.

Part 6 of 6

Linear regression on a graphic display calculator

A line drawn by eye depends on your judgement. Your graphic display calculator can work out a line of best fit from the data themselves. This is the regression line, and the calculator gives its equation in the form $y = a x + b$ : $a$ is the gradient and $b$ is the $y$ -intercept.
Enter the $x$ -values in one list and the $y$ -values in a second list, in the same order, so that each $y$ -value sits next to its own $x$ -value. Check that both lists have the same number of entries.
In the statistics menu, choose the calculation for two variables, then linear regression. The exact key presses depend on your model, so practise them on your own calculator before the exam.
The trap: some calculators offer two forms, $y = a x + b$ and $y = a + b x$ . In the second form $b$ is the gradient, not $a$ . Read which form is on the screen before you copy the numbers.
Write the equation with the values rounded to $3$ significant figures. For the café data in Worked example 1, the screen shows $a = - 2.8333 \dots$ and $b = 113.83 \dots$ , so the regression line is
$y = - 2.83 x + 114$
Compare it with the line drawn by eye in Worked example 2, through $(8, 91)$ and $(24, 46)$ . That line has gradient $- 2.8125$ and meets the $y$ -axis at $113.5$ . The two lines are very close, which is what you expect when the correlation is strong.
To use the equation, substitute. At $13$ °C: $- 2.8333 \dots \times 13 + 113.83 \dots = 77.0$ , so about $77$ hot drinks. Use the calculator's full values in your working and round only the final answer.
The regression line always passes through the mean point. Substituting $\overline{x} = 16$ into the equation with its full values gives $68.5$ , which is $\overline{y}$ . This is a quick check that you entered the lists correctly.
An estimate from the regression equation outside the range of the data is still extrapolation, and still unreliable. Your calculator may also show a value labelled $r$ : this subtopic asks you for the equation and its use.

Worked examples

Example 1. A café records the noon temperature, $x$ °C, and the number of hot drinks sold, $y$ , on eight days. Temperature: $8, 10, 12, 14, 18, 20, 22, 24$ . Hot drinks sold: $92, 84, 82, 71, 65, 59, 48, 47$ , in the same order. (a) Draw a scatter diagram to show the data. (b) Describe the correlation, and say what it shows about the café's sales.

Show the working

Put temperature on the horizontal axis, because it is the variable the owner would use to predict sales.
Choose scales that fit the data. Temperature runs from $8$ to $24$ , so a horizontal axis from $0$ to $25$ , with $2$ cm for every $5$ °C, works. Drinks sold run from $47$ to $92$ , so a vertical axis from $40$ to $100$ , with $2$ cm for every $10$ drinks, works.
Work out each small square. On $2$ mm squared paper, $2$ cm is $10$ small squares. Across, one small square is $5 \div 10 = 0.5$ °C. Up, one small square is $10 \div 10 = 1$ drink.
Plot the first pair, $(8, 92)$ : across to $8$ , then up to $92$ , which is $2$ small squares above the $90$ line. Mark a small cross there.
Plot the other seven pairs in the same way, then check each cross against the table.
graph diagram — not drawn yet
Scatter diagram with horizontal axis 'Noon temperature (°C)' from 0 to 25 and vertical axis 'Hot drinks sold' from 40 to 100, with small crosses at (8, 92), (10, 84), (12, 82), (14, 71), (18, 65), (20, 59), (22, 48) and (24, 47), and no line drawn.
The crosses fall from top left to bottom right and lie close to a straight line: strong negative correlation.
(b) Read the pattern. The points fall from top left to bottom right, so the correlation is negative. They lie close to a straight line, so it is strong.
Now say what it means in context, using 'tends to': on hotter days, the café tends to sell fewer hot drinks.

Answer: (a) Eight crosses at the pairs in the table, with temperature across and drinks sold up. (b) Strong negative correlation: the higher the noon temperature, the fewer hot drinks the café tends to sell.

Example 2. Use the café data from Worked example 1. (a) Calculate the mean point. (b) Draw a line of best fit. (c) Use your line to find an estimate of the number of hot drinks sold on a day when the noon temperature is $13$ °C. (d) Find the gradient of your line and explain what it means. (e) The owner wants to use the line to predict sales on a day when the noon temperature is $45$ °C. Comment on this.

Show the working

(a) Mean temperature: add the eight temperatures and divide by $8$ .
$\overline{x} = \frac{8 + 10 + 12 + 14 + 18 + 20 + 22 + 24}{8} = \frac{128}{8} = 16$
Mean number of drinks: add the eight values and divide by $8$ . So the mean point is $(16, 68.5)$ . Plot it and circle it.
$\overline{y} = \frac{92 + 84 + 82 + 71 + 65 + 59 + 48 + 47}{8} = \frac{548}{8} = 68.5$
(b) Lay the ruler through $(16, 68.5)$ and turn it until the points balance. A good position passes through $(8, 91)$ and $(24, 46)$ . Five points lie just above this line and three just below, and there are points on both sides at each end and in the middle, so the line follows the trend.
Draw the line from $x = 8$ to $x = 24$ , so that it runs across the full data set.
graph diagram — not drawn yet
The café scatter diagram (crosses at (8, 92), (10, 84), (12, 82), (14, 71), (18, 65), (20, 59), (22, 48), (24, 47); temperature 0 to 25 across, drinks 40 to 100 up) with the mean point (16, 68.5) circled, a ruled line of best fit from (8, 91) through (16, 68.5) to (24, 46), and dashed reading lines going up from 13 on the temperature axis to the line and across to about 77 on the drinks axis.
The line passes through the circled mean point and runs across the whole data set. The dashed lines show how the estimate at 13 °C is read.
(c) $13$ °C is inside the data range, so this is interpolation. Go up from $13$ on the temperature axis to the line, then across to the drinks axis. The reading is about $77$ .
(d) Use the two points on the line, which are far apart. Then interpret it: for each $1$ °C rise in temperature, the café sells about $2.8$ fewer hot drinks.
$gradient = \frac{46 - 91}{24 - 8} = \frac{- 45}{16} = - 2.8125$
(e) The data go from $8$ °C to $24$ °C only. $45$ °C is far outside that range, so using the line there is extrapolation. This line meets the $y$ -axis at $91 + 8 \times 2.8125 = 113.5$ . Extending it gives $113.5 - 2.8125 \times 45 = - 13.1$ drinks, which is impossible. That shows the pattern cannot continue that far.

Answer: (a) $(16, 68.5)$ (b) a ruled line through $(16, 68.5)$ from $x = 8$ to $x = 24$ , for example through $(8, 91)$ and $(24, 46)$ (c) about $77$ hot drinks (roughly $75$ to $79$ , depending on your line) (d) about $- 2.8$ : roughly $2.8$ fewer hot drinks for each $1$ °C rise (e) not reliable: $45$ °C is outside the data range of $8$ °C to $24$ °C, and the line even predicts a negative number of drinks.

Example 3. A courier records the distance, $x$ km, and the time taken, $y$ minutes, for eight deliveries. Distance: $2, 4, 5, 7, 8, 10, 12, 15$ . Time: $9, 15, 15, 21, 23, 26, 32, 37$ , in the same order. (a) Use your graphic display calculator to find the equation of the regression line, in the form $y = a x + b$ . (b) Use your equation to find an estimate of the time for a delivery of $9$ km. (c) Explain what the value of $a$ tells you. (d) Comment on using the equation for a delivery of $40$ km.

Show the working

(a) Enter the distances in one list and the times in a second list, in the same order. Both lists have $8$ entries.
Run linear regression in the form $y = a x + b$ . The screen shows $a = 2.148997 \dots$ and $b = 5.326647 \dots$
Round each value to $3$ significant figures and write out the equation, not just the two numbers.
$y = 2.15 x + 5.33$
Check with the mean point: $\overline{x} = 63 \div 8 = 7.875$ and $\overline{y} = 178 \div 8 = 22.25$ . With the full values, $2.148997 \dots \times 7.875 + 5.326647 \dots = 22.25$ . The line passes through the mean point, so the lists went in correctly.
(b) $9$ km is inside the data range of $2$ to $15$ km, so this is interpolation. Substitute $x = 9$ using the full values.
$y = 2.148997 \dots \times 9 + 5.326647 \dots = 24.667 \dots \approx 24.7$
With the rounded equation, $2.15 \times 9 + 5.33 = 24.68$ , which also rounds to $24.7$ . The full values are still the safer habit, because rounding early can change the last figure.
(c) $a$ is the gradient. Each extra kilometre adds about $2.15$ minutes to the time of a delivery.
(d) $40$ km is far outside the data range of $2$ to $15$ km. The equation gives about $91.3$ minutes, but that is extrapolation: there are no long trips in the data, so you cannot tell whether the pattern continues. The estimate is unreliable.

Answer: (a) $y = 2.15 x + 5.33$ (b) $24.7$ minutes (c) each extra kilometre adds about $2.15$ minutes (d) unreliable: $40$ km is outside the data range of $2$ to $15$ km, so it is extrapolation.

Where people go wrong

Mistake: Joining the points of a scatter diagram with a zigzag line, dot to dot.

Instead: Plot crosses only. The only line you draw is a single ruled line of best fit, and only when asked.

The diagram is about the overall pattern of the points. A zigzag through them shows nothing about the trend and cannot be used for estimates.

Mistake: Drawing the line of best fit from the first point to the last point, or forcing it through the origin.

Instead: Draw a ruled line through the mean point $(\overline{x}, \overline{y})$ with the points balanced either side along its whole length.

The end points are only two of the data and may be outliers. The origin is not a data point unless it is in the table, so there is no reason for the trend to pass through it.

Mistake: Calling a pattern with no trend 'negative correlation', or thinking that 'negative' means 'weak'.

Instead: Negative is a direction: one variable tends to decrease as the other increases. No trend at all is zero correlation.

Direction (positive, negative, zero) and strength (strong, weak) are separate descriptions. A negative correlation can be strong.

Mistake: Trusting an estimate from the line, or from the regression equation, for a value far outside the data.

Instead: Say that the estimate is unreliable because it is outside the range of the data, and quote that range.

There are no data there, so there is no evidence that the pattern continues. The café line predicts a negative number of drinks at $45$ °C.

Mistake: Copying the calculator's $a$ as the gradient when the screen shows the form $y = a + b x$ .

Instead: Check which form is on the screen. The gradient is always the number that multiplies $x$ .

In $y = a + b x$ the letters swap jobs: $b$ is the gradient and $a$ is the intercept. Swapping them gives a completely different line.

Check yourself

A scatter diagram shows the age of a car across and its value up. The points fall from top left to bottom right. Write down the type of correlation, and describe the relationship in context.
Answer
Negative correlation. Older cars tend to be worth less.
Describe what a scatter diagram with zero correlation looks like, and say whether you should draw a line of best fit on it.
Answer
The points are scattered with no upward or downward trend. You do not draw a line of best fit, because there is no trend for it to follow.
On a horizontal axis, $10$ small squares represent $5$ cm. How many small squares along from $0$ do you go to plot the value $13.5$ cm?
Answer
One small square is $5 \div 10 = 0.5$ cm, so $13.5 \div 0.5 = 27$ small squares.
The $x$ -values are $2, 4, 6, 8, 10$ and the $y$ -values are $11, 14, 14, 19, 22$ . Calculate the mean point.
Answer
$\overline{x} = 30 \div 5 = 6$ and $\overline{y} = 80 \div 5 = 16$ , so the mean point is $(6, 16)$ .
A line of best fit passes through $(2, 15)$ and $(10, 39)$ . Find its gradient, and use it to find an estimate of $y$ when $x = 6$ .
Answer
Gradient $= (39 - 15) \div (10 - 2) = 24 \div 8 = 3$ . From $x = 2$ to $x = 6$ is $4$ more, so $y = 15 + 4 \times 3 = 27$ .
Data were collected for values of $x$ from $10$ to $50$ . Which estimate from the line of best fit is more reliable, the one at $x = 30$ or the one at $x = 80$ ? Give a reason.
Answer
The one at $x = 30$ . It is inside the range of the data (interpolation). $x = 80$ is outside the range (extrapolation), where the pattern may not continue.
A graphic display calculator gives the regression equation $y = 1.85 x + 12.4$ . Find an estimate of $y$ when $x = 20$ .
Answer
$y = 1.85 \times 20 + 12.4 = 37 + 12.4 = 49.4$ .
A calculator set to the form $y = a + b x$ shows $a = 0.62$ and $b = 3.5$ . Write down the equation of the regression line, and state its gradient.
Answer
$y = 0.62 + 3.5 x$ . The gradient is $3.5$ , the number that multiplies $x$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The table shows the age, in years, and the price, in dollars, of ten second-hand bicycles of the same model. Age (years): 1, 1, 2, 3, 3, 4, 5, 6, 7, 8 Price (dollars): 450, 420, 400, 340, 360, 300, 260, 230, 180, 160 On graph paper, draw a horizontal axis for age from 0 to 10 years, using 1 cm for 1 year, and a vertical axis for price from 0 to 500 dollars, using 1 cm for 50 dollars.

(a) Plot the ten points on your grid. 2 marks No calculator

Mark scheme

B2 all ten points correct, each within half a small square. B1 for 8 or 9 points correct

(b) State the type of correlation shown by your scatter diagram. 1 mark No calculator

Mark scheme

B1 negative (allow strong negative)

(c) Calculate the mean age and the mean price. 2 marks No calculator

Mark scheme

B1 mean age 4 (years)
B1 mean price 310 (dollars)

(d) Draw a line of best fit on your scatter diagram. 2 marks No calculator

Mark scheme

B1 a single ruled line passing through their mean point, (4, 310) if correct
B1 the line covers the full data set, from age 1 to age 8, with the points roughly evenly either side, e.g. passing between (1, 420) and (1, 450) and between (8, 130) and (8, 170)

(e) Find an estimate of the price of a bicycle of this model that is $4 \frac{1}{2}$ years old. Use your line of best fit. 1 mark No calculator

Mark scheme

B1 answer in the range 280 to 300 dollars, or FT a correct reading from their ruled line

(f) Explain why your line of best fit should not be used to find the price of a bicycle of this model that is $15$ years old. 1 mark No calculator

Mark scheme

B1 15 years is outside the range of the data (1 to 8 years), so the trend may not continue oe, e.g. extending the line gives a negative price

Question 2. A plant nursery measures eight young trees of the same type. The table shows each tree's age, $t$ months, and its height, $h$ cm. $t$ : 3, 5, 6, 8, 10, 12, 14, 16 $h$ : 20, 31, 32, 42, 49, 58, 65, 73

(a) Find the equation of the regression line, in the form $h = a t + b$ . 2 marks

Mark scheme

B2 h = 4.02t + 9.09, a and b awrt. B1 for a = awrt 4.02 or b = awrt 9.09 in the correct position

(b) Use your equation to find an estimate of the height of a tree of this type that is $11$ months old. 2 marks

Mark scheme

M1 4.02 × 11 + 9.09 oe, substituting 11 into their equation
A1 awrt 53.3 (cm)

(c) The owner uses the equation to predict the height of a tree of this type at $60$ months old. Comment on the reliability of this prediction. 2 marks

Mark scheme

B1 not reliable oe
B1 dep on the first B1: 60 months is outside the range of the data (3 to 16 months), so the trend may not continue oe

(d) Explain what the value of $a$ in your equation tells you about the trees. 1 mark

Mark scheme

B1 the height increases by about 4 cm (4.02 cm) for each month of age oe

Question 3. Each pair of variables below is recorded for a set of items. A: the number of pages in a paperback novel and the mass of the novel B: the time a candle has been burning and the height of the candle, measured every 10 minutes for one candle C: an adult's height and the day of the month on which the adult was born

(a) Write down the type of correlation you would expect for each of A, B and C. 2 marks No calculator

Mark scheme

B2 A positive, B negative, C zero (allow none or no correlation). B1 for two correct

(b) Explain what is meant by zero correlation. 1 mark No calculator

Mark scheme

B1 there is no relationship between the two variables: as one increases, the other does not tend to increase or decrease oe

How this is examined

Drawing a line of best fit through the mean point: a Calculate for the mean of each variable, followed by a Draw for the line on the scatter diagram you have plotted Both means correct, for example $\overline{x} = 128 \div 8 = 16$ and $\overline{y} = 548 \div 8 = 68.5$ . Then one ruled straight line that passes through $(\overline{x}, \overline{y})$ , runs across the full data set from the smallest $x$ -value to the largest, and has the points roughly balanced either side along its whole length. A line joining the first and last points, a line forced through the origin or a freehand line does not meet this description.
Finding the regression equation on the calculator and using it: a calculator-paper Find asking for the equation in a stated form such as $y = a x + b$ , followed by a Find that substitutes a given value The equation written out in the form asked for, with $a$ and $b$ to $3$ significant figures, such as $y = 2.15 x + 5.33$ , not just two numbers. For the estimate, the substitution shown, such as $2.15 \times 9 + 5.33$ , and the answer to $3$ significant figures. If the screen uses $y = a + b x$ , the gradient is $b$ .
Estimating from the line, and saying whether to trust the estimate: a one-mark Find that reads a value from your own line, then a one- or two-mark Explain or Comment about a value outside the data A reading taken from your own line, with the reading lines drawn. For reliability: say whether the value is inside or outside the range of the data and quote the range, for example 'unreliable, because $45$ °C is outside the data range of $8$ °C to $24$ °C, so the trend may not continue'. 'It is not accurate' with no reason is not a Comment.
Plotting the points accurately: a two-mark Plot on a grid with the scales given, from a table of pairs Each point marked with a small cross whose centre is on the correct position, to within half a small square. Work out what one small square is worth on each axis before you start, and tick off each pair in the table as you plot it.
Naming the correlation and saying what it means: a one-mark State or Write down for the type of correlation, which may be followed by a Describe of the relationship in context The word positive, negative or zero for the type. For the relationship, a sentence in context that gives the direction, such as 'the older the bicycle, the lower its price tends to be'.

Words to know

Scatter diagram: A graph of paired data, with one point plotted for each pair of values.
Correlation: The way two variables change together, judged from the overall pattern of the points on a scatter diagram.
Positive correlation: As one variable increases, the other tends to increase. The points rise from bottom left to top right.
Negative correlation: As one variable increases, the other tends to decrease. The points fall from top left to bottom right.
Zero correlation: No relationship: the points show no upward or downward trend. Also called no correlation.
Mean point: The point $(\overline{x}, \overline{y})$ , made from the mean of the $x$ -values and the mean of the $y$ -values. A line of best fit passes through it.
Line of best fit: One straight ruled line that follows the trend of the points, passes through the mean point and runs across the full data set.
Outlier: A point that lies well away from the pattern of the other points.
Interpolation: Making an estimate inside the range of the data. It is reasonably reliable.
Extrapolation: Making an estimate outside the range of the data. It is unreliable, because the pattern may not continue.
Regression line: The line of best fit that a graphic display calculator works out from the data, given as an equation such as $y = a x + b$ .

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